Probability Problem

Ok, here's the problem i've been having some trouble with:

Ann and Bob have agreed to meet for dinner "around four". The number of minutes after 4:00 that Ann will arrive is approximately normally distributed with mean 10 and standard deviation 3.1, and the number of minutes after 4:00 that Bob will arrive is approximately normally distributed with mean 20 and standard deviation 4.2.

If the longest each will wait is 10 minutes, find the probability that they'll meet?

I know the answer will be around the 1 range but i don't really know how to show the work for it, at least for the work that i showed for it, i'm not confident about it. I'll be very appreciated if someone can help me out. Thank You in advance.

Ann and Bob have agreed to meet for dinner "around four". The number of minutes after 4:00 that Ann will arrive is approximately normally distributed with mean 10 and standard deviation 3.1, and the number of minutes after 4:00 that Bob will arrive is approximately normally distributed with mean 20 and standard deviation 4.2.

If the longest each will wait is 10 minutes, find the probability that they'll meet?

I know the answer will be around the 1 range but i don't really know how to show the work for it, at least for the work that i showed for it, i'm not confident about it. I'll be very appreciated if someone can help me out. Thank You in advance.

Consider the difference D in arrival times, this will be normally distributed with
mean 10 minutes, and sd sqrt(3.1^2+4.2^2). Then you want: p(-10<D<10).